# Cengel Solutions

**Topics:**Fluid dynamics, Velocity, Fluid mechanics

**Pages:**190 (22184 words)

**Published:**November 30, 2012

Solutions Manual for

Fluid Mechanics: Fundamentals and Applications

by Çengel & Cimbala

CHAPTER 4

FLUID KINEMATICS

PROPRIETARY AND CONFIDENTIAL

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4-1

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 4 Fluid Kinematics

Introductory Problems

4-1C

Solution

We are to define and explain kinematics and fluid kinematics.

Analysis

Kinematics means the study of motion. Fluid kinematics is the study of how fluids flow and how to describe fluid motion. Fluid kinematics deals with describing the motion of fluids without considering (or even understanding) the forces and moments that cause the motion. Discussion

Fluid kinematics deals with such things as describing how a fluid particle translates, distorts, and rotates, and how to visualize flow fields.

4-2

Solution

We are to write an equation for centerline speed through a nozzle, given that the flow speed increases parabolically.

Assumptions

1 The flow is steady. 2 The flow is axisymmetric. 3 The water is incompressible.

Analysis

A general equation for a parabola in the x direction is

u = a + b ( x − c)

General parabolic equation:

2

(1)

We have two boundary conditions, namely at x = 0, u = uentrance and at x = L, u = uexit. By inspection, Eq. 1 is satisfied by setting c = 0, a = uentrance and b = (uexit - uentrance)/L2. Thus, Eq. 1 becomes u = uentrance +

Parabolic speed:

( uexit − uentrance )

L2

x2

(2)

Discussion

You can verify Eq. 2 by plugging in x = 0 and x = L.

4-3

Solution

location.

For a given velocity field we are to find out if there is a stagnation point. If so, we are to calculate its

Assumptions

1 The flow is steady. 2 The flow is two-dimensional in the x-y plane.

Analysis

The velocity field is

V = ( u , v ) = ( 0.5 + 1.2 x ) i + ( −2.0 − 1.2 y ) j

(1)

At a stagnation point, both u and v must equal zero. At any point (x,y) in the flow field, the velocity components u and v are obtained from Eq. 1,

Velocity components:

u = 0.5 + 1.2 x

v = −2.0 − 1.2 y

(2)

x = −0.4167

y = −1.667

(3)

Setting these to zero yields

Stagnation point:

0 = 0.5 + 1.2 x

0 = −2.0 − 1.2 y

So, yes there is a stagnation point; its location is x = -0.417, y = -1.67 (to 3 digits). Discussion

If the flow were three-dimensional, we would have to set w = 0 as well to determine the location of the stagnation point. In some flow fields there is more than one stagnation point.

4-2

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 4 Fluid Kinematics

4-4

Solution

location.

For a given velocity field we are to find out if there is a stagnation point. If so, we are to calculate its

Assumptions

1 The flow is steady. 2 The flow is two-dimensional in the x-y...

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